Akangaroo can jump over an object 2.50 m high. (a) considering just its vertical motion, calculate its vertical speed when it leaves the ground. (b) how long a time is it in the air?

a) Vertical speed when it leaves the ground is 7 m/sb) Kangaroo is in air for 1.43 seconds.

Explanation:

a) We have equation of motion v² = u² + 2as

Initial velocity, u = ?

Acceleration, a = -9.81 m/s²  

Final velocity, v = 0 m/s  

Displacement, s = 2.50 m

Substituting  

v² = u² + 2as

0² = u² + 2 x -9.81 x 2.50

u = 7 m /s

Vertical speed when it leaves the ground is 7 m/s

b)  We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 7 m/s

        Acceleration, a = -9.81 m/s²  

        Displacement, s = 0 m   --- Time is calculating for returning to ground    

     Substituting

                      s = ut + 0.5 at²

                      0 = 7 x t + 0.5 x -9.81 x t²

                      t = 1.43 seconds

Kangaroo is in air for 1.43 seconds.

100 pounds is 100 pounds no matter the object or item being weighed

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward